- #1
sponsoraw
- 42
- 1
Homework Statement
Homework Equations
The Attempt at a Solution
Not sure if I'm on the right track here at all.
p=40*10-6*6*10-3*290*109/1=69600 Pa=69.6 kPa
Simple Calculation for Pressure Vessel Strain | 69.6 kPa Solution
Click For Summary
Discussion Overview
The discussion revolves around the calculation of strain in a pressure vessel, specifically focusing on the hoop strain and the associated stresses. Participants explore the application of Hooke's law and the relationships between hoop and axial stresses in the context of multiaxial loading.
Discussion Character
- Homework-related
- Technical explanation
- Debate/contested
Main Points Raised
- One participant presents a calculation for pressure resulting in 69.6 kPa but expresses uncertainty about the approach taken.
- Another participant argues that the initial method used to calculate hoop stress is incorrect due to the biaxial state of stress in the vessel.
- A participant questions whether to use axial stress instead of hoop stress, citing the relationship between the two stresses.
- Further clarification is sought regarding the correct equations for hoop and axial strains in terms of the respective stresses.
- One participant confirms a formula for hoop strain and inquires about the Poisson ratio, indicating a need for additional information.
- A later reply acknowledges the need for the Poisson ratio and provides a value, suggesting that the calculation can proceed once the pressure is isolated in the formula.
Areas of Agreement / Disagreement
Participants express differing views on the correct approach to calculating stresses and strains in the pressure vessel, indicating that multiple competing views remain unresolved.
Contextual Notes
Participants reference the need for specific material properties, such as the Poisson ratio, which was initially omitted. The discussion also highlights the complexity of applying Hooke's law in a multiaxial context, suggesting that assumptions about stress relationships may need further exploration.
Not sure if I'm on the right track here at all.
p=40*10-6*6*10-3*290*109/1=69600 Pa=69.6 kPa
- #2
- 23,740
- 5,940
You used the equation ε=σ/E to calculate the hoop stress, but this approach is not correct because the state of stress is biaxial. You need to calculate the axial stress in terms of the pressure, and then use Hooke's law in its tensorial form to determine the pressure.
Chet
Chet
- #3
sponsoraw
- 42
- 1
Chet, I might not understand it correct. The reason why I've used the hoop stress is that in the question the strain gauges are installed on the vessel to measure the hoop stress as hoop stress = 2x axial stress. Should I use the axial stress instead, why?
- #4
- 23,740
- 5,940
No. The problem is with the equation you used to get the hoop strain. In terms of the (unknown) pressure, your equation for the hoop stress is correct. You indicated that the axial stress is half the hoop stress. So, in terms of the (unknown) pressure p, what is the axial stress? What are the Hooke's law equations for the hoop stain and the axial strain in terms of the hoop stress and the axial stress (both equations involve both stresses)?
Chet
Chet
- #5
sponsoraw
- 42
- 1
Chet, looks like I may need a bit more help.
- #6
- 23,740
- 5,940
Have you learned about the general form(s) of the Hooke's law relationship for multiaxial loading?sponsoraw said:
Chet
- #7
sponsoraw
- 42
- 1
Will the hoop strain for a closed-end cylindrical pressure vessel be ε=(p*ri/t*E)*(1-v/2)? How do I get v?
- #8
- 23,740
- 5,940
Yes. Nice job. You need to look up the Poisson ratio for the pipe material. They give you the Young's modulus, but not the Poisson ratio?sponsoraw said:
Chet
- #9
sponsoraw
- 42
- 1
Chet, thanks for you help with this. I've contacted the tutor and he accidentally forgot to include the Poisons ratio, it's v=0.3. It should be straight forward now, just need to make p the subject of the formula. I got p=(εEt)/[r(1-0.5v)].
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